Adding and subtracting still the suma024+12⁢∑i=1n(ai2+bi2)superscriptsubscripta02412superscriptsubscripti1nsuperscriptsubscriptai2superscriptsubscriptbi2\frac{a_{0}^{2}}{4}+\frac{1}{2}\sum_{{i=1}}^{n}(a_{i}^{2}+b_{i}^{2}) yields finally the form

The three first addends of this sum do not depend on the choice of the quantities αisubscriptαi\alpha_{i} and βisubscriptβi\beta_{i}. The other addends are non-negative, and their sum is minimal, equal 0, when